\(a.\)

\(-\dfrac{2}{9}+\dfrac{7}{-9}=\dfrac{-2}{9}-\dfrac{7}{9}=\dfrac{-2-7}{9}=\dfrac{-9}{9}=-1\)

\(b.\)

\(\dfrac{1}{-6}+\dfrac{13}{-15}=-\dfrac{1}{6}-\dfrac{13}{15}=\dfrac{\left(-1\right)\cdot5-13\cdot2}{30}=\dfrac{-5-26}{30}=\dfrac{-31}{30}\)

\(c.\)

\(\dfrac{5}{-6}+\dfrac{-5}{12}+\dfrac{7}{18}=\dfrac{-5}{6}-\dfrac{5}{12}+\dfrac{7}{18}=\dfrac{\left(-5\right)\cdot6-5\cdot3+7\cdot2}{36}=\dfrac{-31}{36}\)

a) \(\dfrac{-2}{9}+\dfrac{7}{-9}=\dfrac{-2+-7}{9}=-1\) 

b)\(\dfrac{1}{-6}+\dfrac{13}{-15}=\dfrac{-5}{30}+\dfrac{26}{30}=\dfrac{-5+26}{30}=\dfrac{7}{10}\) 

c) \(\dfrac{5}{-6}+\dfrac{-5}{12}+\dfrac{7}{18}=\dfrac{-30}{36}+\dfrac{-15}{36}+\dfrac{18}{36}=\dfrac{-30+-15+18}{36}=\dfrac{-3}{4}\)

2:

a: \(2^2\cdot3-4=4\cdot3-4=12-4=8\)

b: \(16-2^3\cdot2=16-2^4=16-16=0\)

c: \(4^2-4\cdot2=16-8=8\)

d: \(3^3-2\cdot3^2=27-2\cdot9=27-18=9\)

e: \(7^2-9\cdot2^2=49-9\cdot4=49-36=13\)

f: \(2^2\cdot3+4^2=4\cdot3+16=12+16=28\)

Bài 1:

a: \(13+21\cdot5-\left(198:11-8\right)\)

\(=13+105-18+8\)

=21+87

=108

b: \(272:16-5+4\cdot\left(30-5-255:17\right)\)

\(=17-5+4\cdot\left(30-5-15\right)\)

\(=12+4\cdot10=12+40=52\)

c: \(15\cdot24-14\cdot5\cdot\left(145:5-27\right)\)

\(=360-70\left(29-27\right)\)

=360-140

=220

d: \(18\cdot3-18\cdot2+3\cdot\left(\dfrac{51}{17}\right)\)

\(=18\left(3-2\right)+3\cdot3\)

=18+9

=27

e: \(64+115+36-25\cdot8\cdot6\cdot2^2\cdot3+4^2\)

\(=100+115-200\cdot6\cdot4\cdot3+16\)

\(=231-14400=-14169\)

f: \(15\cdot8-\left(17-30+83\right)-144:6\)

\(=120-24-\left(100-30\right)\)

=96-70

=26

   \(\dfrac{3}{17}\) .\(\dfrac{6}{29}\) - \(\dfrac{3}{17}\).\(\dfrac{35}{29}\) + 2022\(\dfrac{3}{17}\)

= (\(\dfrac{3}{17}\).\(\dfrac{6}{29}\) - \(\dfrac{3}{17}\).\(\dfrac{35}{29}\)) + 2022 + \(\dfrac{3}{17}\)

= \(\dfrac{3}{17}\).(\(\dfrac{6}{29}\) - \(\dfrac{35}{29}\)) + 2022 + \(\dfrac{3}{17}\)

= \(\dfrac{3}{17}\).(-1) + 2022 + \(\dfrac{3}{17}\)

 = (- \(\dfrac{3}{17}\) + \(\dfrac{3}{17}\)) + 2022

= 0 + 2022

= 2022 

b; \(\dfrac{5}{6}\) - (\(\dfrac{1}{3}\) + \(\dfrac{1}{2}\)).20%

= \(\dfrac{5}{6}\) - \(\dfrac{5}{6}\).\(\dfrac{1}{5}\)

= \(\dfrac{5}{6}-\dfrac{1}{6}\)

= \(\dfrac{4}{6}\) 

= \(\dfrac{2}{3}\)

Tính:

A.17 phút 21 giây+22 phút 15 giây= 39 phút 36 giây

B.25 ngày 6 giờ-4 ngày 9 giờ= 20 ngày 21 giờ

C.14 phút 42 giây*2= 28 phút 84 giây

D.5 phút 18 giây:2= 2 phút 39 giây

\(1,\\ c,\dfrac{17}{6}-\dfrac{3}{7}-\dfrac{5}{6}=\dfrac{17\times7-3\times6-5\times7}{6\times7}=\dfrac{119-18-35}{42}=\dfrac{66}{42}=\dfrac{11}{7}\\ 2,\\ a,\dfrac{6}{9}+\dfrac{3}{15}+\dfrac{7}{15}\\ =\dfrac{6:3}{9:3}+\dfrac{7+3}{15}\\ =\dfrac{2}{3}+\dfrac{10}{15}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{2+2}{3}=\dfrac{4}{3}\)

\(b,\dfrac{7}{2}+\dfrac{29}{12}-\dfrac{11}{12}\\ =\dfrac{7}{2}+\left(\dfrac{29}{12}-\dfrac{11}{12}\right)\\ =\dfrac{7}{2}+\dfrac{29-11}{12}\\ =\dfrac{7}{2}+\dfrac{18}{12}\\ =\dfrac{7}{2}+\dfrac{18:6}{12:6}\\ =\dfrac{7}{2}+\dfrac{3}{2}\\ =\dfrac{7+3}{2}=\dfrac{10}{2}=5\)

\(c,\dfrac{26}{25}-\dfrac{3}{5}-\dfrac{2}{5}\\ =\dfrac{26}{25}-\dfrac{3\times5}{5\times5}-\dfrac{2\times5}{5\times5}\\ =\dfrac{26-15-10}{25}\\ =\dfrac{1}{25}\)

Bài 1:
\(\dfrac{17}{6}-\dfrac{3}{7}-\dfrac{5}{6}=\left(\dfrac{17}{6}-\dfrac{5}{6}\right)-\dfrac{3}{7}=2-\dfrac{3}{7}=\dfrac{14}{7}-\dfrac{3}{7}=\dfrac{11}{7}\)
Bài 2:
a/\(\dfrac{6}{9}+\dfrac{3}{15}+\dfrac{7}{15}\)
\(=\dfrac{2}{3}+\left(\dfrac{3}{15}+\dfrac{7}{15}\right)\)
\(=\dfrac{2}{3}+\dfrac{2}{3}\)
\(=\dfrac{4}{3}\)
b/\(\dfrac{7}{2}+\dfrac{29}{12}-\dfrac{11}{12}\)
\(=\dfrac{7}{2}+\left(\dfrac{29}{12}-\dfrac{11}{12}\right)\)
\(=\dfrac{7}{2}+\dfrac{3}{2}\)
\(=5\)
c/\(\dfrac{26}{25}-\dfrac{3}{5}-\dfrac{2}{5}\)
\(=\dfrac{26}{25}-\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\)
\(=\dfrac{26}{25}-1\)
\(=\dfrac{26}{25}-\dfrac{25}{25}\)
\(=\dfrac{1}{25}\)
#Đang Bận Thở

a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)

=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)

=\(\left(-1\right)+1+\dfrac{-2}{11}\)

=\(\dfrac{-2}{11}\)

b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)

=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)

=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)

=\(0+0+\dfrac{7}{17}\)

=\(\dfrac{7}{17}\)

c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)

A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)

A=\(35-5\dfrac{7}{32}\)

A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)

d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)

C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)

C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)

C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)

a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`

`=-5/9+8/15-2/11-4/9+7/15`

`=(-5/9-4/9)+(8/15+7/15)-2/11`

`=-9/9+15/15-2/11`

`=-1+1-2/11`

`=-2/11`

b, `((-5)/12+6/11)+(7/17+5/11+5/12)`

`=-5/12+6/11+7/17+5/11+5/12`

`=(-5/12+5/12)+(6/11+5/11)+7/17`

`=0+11/11+7/17`

`=1+7/17`

`=17/17+7/17`

`=24/17`

c, `A=49 8/23 - (5 7/32 + 14 8/23)`

`A=49 8/23 - 5 7/32 - 14 8/23`

`A=(49 8/23 - 14 8/23)-5 7/32`

`A=35 - 167/32`

`A=953/32`

d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`

`C=-3/7 . 5/9-4/9 . 3/7+17/7`

`C=-3/7.(5/9+4/9)+17/7`

`C=-3/7 . 1+17/7`

`C=2`

a) Ta có: \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{-4}{9}+\dfrac{7}{15}\)

\(=\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)-\dfrac{2}{11}\)

\(=-1+1-\dfrac{2}{11}\)

\(=-\dfrac{2}{11}\)

b) Ta có: \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{11}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)

\(=\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{7}{11}+\dfrac{5}{11}\right)\)

\(=\dfrac{18}{11}\)

c) Ta có: \(A=49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

\(=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}\)

\(=30-\dfrac{7}{32}=\dfrac{953}{32}\)

a, 1. \(\dfrac{2}{17}\)

2. \(22\)

3. 4

b,

1. \(\dfrac{8}{5}\)

2. \(\dfrac{176}{5}\)

3. \(\dfrac{28}{5}\)

4. \(\dfrac{3}{10}\)

5. \(\dfrac{163}{5}\)

6. \(\dfrac{9}{20}\)

a) \(\dfrac{12}{17}:6=\dfrac{12}{17}\cdot\dfrac{1}{6}=\dfrac{2}{17}\)

a) \(16:\dfrac{8}{11}=16\cdot\dfrac{11}{8}=22\)

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

Giải:

a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)

\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\) 

\(=\dfrac{107}{220}+\dfrac{43}{20}\)

\(=\dfrac{29}{11}\)

b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\) 

\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\) 

\(=\dfrac{31}{7}:\dfrac{31}{5}\) 

\(=\dfrac{5}{7}\) 

c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\) 

\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\) 

\(=\dfrac{60}{23}:\dfrac{5}{23}\) 

\(=12\)